Four birds and a tetrahedron

4 birds are located at the vertices of a regular tetrahedron with unit side-length. Let us pick the following coordinates as the vertices of the tetrhedron. $$A(\frac{1}{2}, 0, -\frac{1}{2 \sqrt{2}}), \quad B(-\frac{1}{2}, 0, -\frac{1}{2\sqrt{2}}), \quad C(0, \frac{1}{2}, \frac{1}{2\sqrt{2}}), \quad D(0, -\frac{1}{2}, \frac{1}{2\sqrt{2}}) $$

Every bird starts flying towards their consecutive bird (A towards B, B towards C, and so on), until they eventually collide. The question is, for how long do they need to fly until the collision?

The animation above (rotateable!) shows the motion of the four birds. From there, it is obvious that the birds will not stay in a tetrahedronal shape. Fortunately, there is a perspective from which the birds will stay in a consistent configuration: when viewed from the axis (1, 1, 0), the birds are configured in a square. Also, the velocities of the 4 birds possess a rotational symmetry about this axis, thanks to which the birds will always remain in a square when viewed from the axis.

It might seem that this problem now reduces down to the well-known problem of four turtles/bugs in a square. However, this is not the case as the birds also fly in the direction of the axis.

Let us label the distance of bird A from the centroid along the axis as \(h\). This distance is therefore \(-h\) for bird B, \(h\) for C and \(-h\) for D. All birds are at the radius r from the axis and consecutive birds form an angle \(\frac{\pi}{2}\) with the axis.

Consecutive birds, say A and B, are separated by a distance $$| \vec{AB}| =\sqrt{2r^2+4h^2}$$ between them. The magnitude of bird A's velocity along the axis is $$\frac{dh}{dt}=-\frac{2h}{\sqrt{2r^2+4h^2}}v$$ and it's radial velocity is $$\frac{dr}{dt}=-\frac{r}{\sqrt{2r^2+4h^2}}v.$$

Putting these expressions together, we get $$\frac{dh}{dr}=\frac{2h}{r} \quad \text{ or } \quad \ln h = \ln r + C.$$ Knowing that initially \(r_0=\frac{1}{2}\) and \(h_0=\frac{\sqrt{2}}{2}\), we get $$h=2\sqrt{2}r^2.$$

Knowing this is enough to find the distance which the bird flies before the collisions. In a square configuration, the radial velocity of any bird is always equal to its tangential velocity, that is why in a time interval \(dt\) the bird will traverse a path \(\sqrt{2(dr)^2+(dh)^2}=\sqrt{2}\sqrt{1+2r^2}\,dr\). Hence, the total traversed path is $$\int_0^{1/2} \sqrt{2}\sqrt{1+16r^2}\,dr= \frac{\text{arcsinh}(1)+\sqrt{2}}{2^{\frac{3}{2}}}$$